The following reaction has an equilibrium constant (Keq) of 160.
2 NO2(g) <--> 2 NO(g) + O2(g)
What will be the reaction quotient (Q) and in which direction will the reaction proceed if the partial pressure of NO2 is 5.0*10^-4 atm, NO is 0.080 atm, and O2 is 0.020 atm?
Work:
Keq= [NO]^2 [O2]/[NO2]
Keq=[0.080]^2 [.020]/ 5.0*10^-4
Keq=1.28*10^-4/5.0*10^-4
Keq=.256 which is less than 1 meaning that the reactants are favored moving the reaction towards the left.
I am not sure if .256 is also the reaction quotient or if it's a totally different process to find it.
Nevermind I figured it out
In this case, you have correctly calculated the equilibrium constant, Keq, which represents the ratio of the product concentrations to reactant concentrations at equilibrium. However, the value of Keq does not directly provide information about the reaction quotient, Q, at any given point in the reaction.
To determine the reaction quotient, Q, we use the same equation as for Keq, but we evaluate the concentrations or partial pressures of the reactants and products at any given time rather than at equilibrium. This allows us to assess the position of the reaction relative to equilibrium.
In this case, you are given partial pressures for the species involved in the reaction. The reaction quotient, Q, is expressed as:
Q = ([NO]^2 [O2])/[NO2]
Using the given values, we can calculate Q:
Q = (0.080^2 * 0.020)/(5.0 * 10^-4)
Q ≈ 2.56
Comparing the value of Q to the equilibrium constant, Keq, we can determine the direction in which the reaction will proceed. If Q is less than Keq, then the reactants are favored, and the reaction will proceed in the forward direction (to the right). If Q is greater than Keq, then the products are favored, and the reaction will shift in the reverse direction (to the left).
In this case, Q = 2.56, which is greater than the equilibrium constant Keq = 160. Therefore, the reaction will proceed in the reverse direction, meaning that the reactants (NO and O2) will react to form more NO2.